sketching reciprocal function (1 Viewer)

mathsbrain

Member
Joined
Jul 16, 2012
Messages
161
Gender
Male
HSC
N/A
Hi,

1) Just wondering what happens when you reciprocate a vertical asymptote.
For example if you are given the graph y=1/x and asked to reciprocate it, will the origin be an open circle or closed circle.

2) If you are NOT given the equation of y=1/x but given something that looks similar, how can you tell if you are going to get a straight line or a curve after reciprocation?

Thanks!
 
Last edited:

blyatman

Well-Known Member
Joined
Oct 11, 2018
Messages
539
Gender
Undisclosed
HSC
N/A
1) Generally, it'll be a closed circle. It behaves similar to say, y = x/x, which is equal to 1 as long as x =/= 0. So in a similar fashion, the reciprocal of 1/x is 1/(1/x), which is equal to x as long as x =/= 0.

Here's a relatable post: https://math.stackexchange.com/questions/213928/what-is-cot-pi-2

2) Doesn't really matter, as the examiner will only care about the general behaviour of the curve (i.e. whether it's increasing or decreasing). Personally, I'd plot them as curves since we shouldn't assume the original graph to be a hyperbola.
 
Last edited:

mathsbrain

Member
Joined
Jul 16, 2012
Messages
161
Gender
Male
HSC
N/A
Thanks, but just a little confused regarding point 1), you mentioned x=/=0, so why is that a closed circle then?

1) Generally, it'll be a closed circle. It behaves similar to say, y = x/x, which is equal to 1 as long as x =/= 0. So in a similar fashion, the reciprocal of 1/x is 1/(1/x), which is equal to x as long as x =/= 0.

Here's a relatable post: https://math.stackexchange.com/questions/213928/what-is-cot-pi-2

2) Doesn't really matter, as the examiner will only care about the general behaviour of the curve (i.e. whether it's increasing or decreasing). Personally, I'd plot them as curves since we shouldn't assume the original graph to be a hyperbola.
 

Users Who Are Viewing This Thread (Users: 0, Guests: 1)

Top